8. Hypothesis Testing

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1 FE661 - Statistical Methods for Financial Engineering 8. Hypothesis Testing Jitkomut Songsiri introduction Wald test likelihood-based tests significance test for linear regression 8-1

2 Introduction elements of statistical tests null hypothesis, alternative hypothesis test statistics rejection region type of errors: type I and type II errors confidence intervals, p-values examples of hypothesis tests: hypothesis tests for the mean, and for comparing the means hypothesis tests for the variance, and for comparing variances Hypothesis Testing 8-2

3 Testing procedures a test consists of providing a statement of the hypotheses (H 0 (null) and H 1 (alternative)) giving a rule that dictates if H 0 should be rejected or not the decision rule involves a test statistic calculated on observed data the Neyman-Pearson methodology partitions the sample space into two regions the set of values of the test statistic for which: the null hypothesis is rejected we fail to reject the null hypothesis rejection region acceptance region Hypothesis Testing 8-3

4 Test errors since a test statistic is random, the same test can lead to different conclusions type I error: the test leads to reject H 0 when it is true type II error: the test fails to reject H 0 when it is false; sometimes called false alarm probabilities of the errors: let β be the probability of type II error the size of a test is the probability of a type I error and denoted by α the power of a test is the probability of rejecting a false H 0 or (1 β) α is known as significance level and typically controlled by an analyst for a given α, we would like β to be as small as possible Hypothesis Testing 8-4

5 Some common tests normal test t-test F -test Chi-square test e.g. a test is called a t-test if the test statistic follows t-distribution two approaches of hypothesis test critical value approach p-value approach Hypothesis Testing 8-5

6 Critical value approach Definition: the critical value (associated with a significance level α) is the value of the known distribution of the test statistic such that the probability of type I error is α steps involved this test 1. define the null and alternative hypotheses. 2. assume the null hypothesis is true and calculate the value of the test statistic 3. set a small significance level (typically α = 0.01, 0.05, or 0.10) and determine the corresponding critical value 4. compare the test statistic to the critical value condition decision the test statistic is more extreme than the critical value reject H 0 the test statistic is less extreme than the critical value accept H 0 Hypothesis Testing 8-6

7 example: hypothesis test on the population mean samples N = 15, α = 0.05 the test statistic is t = x µ s/ N and has t-distribution with N 1 df test H 0 H 1 critical value reject H 0 if right-tail µ = 3 µ > 3 t α,n 1 t t α,n 1 left-tail µ = 3 µ < 3 t α,n 1 t t α,n 1 two-tail µ = 3 µ 3 t α/2,n 1, t α/2,n 1 t t α/2,n 1 or t t α/2,n pdf of t-distribution pdf of t-distribution pdf of t-distribution Hypothesis Testing 8-7

8 p-value approach Definition: the p-value is the probability that we observe a more extreme test statistic in the direction of H 1 steps involved this test 1. define the null and alternative hypotheses. 2. assume the null hypothesis is true and calculate the value of the test statistic 3. calculate the p-value using the known distribution of the test statistic 4. set a significance level α (small value such as 0.01, 0.05) 5. compare the p-value to α condition decision p-value α reject H 0 p-value α accept H 0 Hypothesis Testing 8-8

9 example: hypothesis test on the population mean (same as on page 8-7) samples N = 15, α = 0.01 (have only a 1% chance of making a Type I error) suppose the test statistic (calculated from data) is t = test H 0 H 1 p-value expression p-value right-tail µ = 3 µ > 3 P (t 14 2) left-tail µ = 3 µ < 3 P (t 14 2) two-tail µ = 3 µ 3 P (t 14 2) + P (t 14 2) Right tail Left tail Two tails pdf of t-distribution pdf of t-distribution pdf of t-distribution area A area B = area A + area B right-tail/left-tail tests: reject H 0, two-tail test: accept H 0 Hypothesis Testing 8-9

10 the two approaches assume H 0 were true and determine p-value the probability of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed critical value whether or not the observed test statistic is more extreme than would be expected (called critical value) the null hypothesis is rejected if p-value p value α critical value test statistic critical value Hypothesis Testing 8-10

11 Hypothesis testing in this chapter, we discuss about the following tests Wald test likelihood ratio test Lagrange multiplier (or score) test Hypothesis Testing 8-11

12 Wald test requires estimation of the unrestricted model linear hypotheses in linear models some Wald test statistics examples Hypothesis Testing 8-12

13 Linear hypotheses in linear models a generalization of tests for linear strictions in the linear regression model null and alternative hypotheses for a two-sided test of linear restrictions on the regression parameters in the model: y = Xβ + u are H 0 : Rβ b = 0 H 1 : Rβ b 0 where R R m n of full rank m, β R n and m n for example, one can test β 1 = 1 and β 2 β 3 = 2 the Wald test of Rβ b = 0 is a test of closeness to zero of the sample analogue R ˆβ b where ˆβ is the unrestricted OLS estimator Hypothesis Testing 8-13

14 assumption: suppose u N (0, σ 2 I) then ˆβ N (β, σ 2 (X T X) 1 ) R ˆβ b N (0, σ 2 R(X T X) 1 R T ) under H 0 where Rβ b = 0 define û = y X ˆβ ls, RSS = N N û 2 i, s 2 = RSS/(N n) = (N n) 1 i=1 i=1 û 2 i Facts: s 2 is an unbiased estimate for σ 2 (N n)s 2 /σ 2 χ 2 (N n) Hypothesis Testing 8-14

15 Some Wald test statistics on linear models known variance σ 2 (cannot be calculated in practice) W 1 = (R ˆβ b) T (σ 2 R(X T X) 1 R T ) 1 (R ˆβ b) χ 2 (m) under H 0 replace σ 2 by any consistent estimate s 2 (not necessarily s 2 on page 8-14) W 2 = (R ˆβ b) T (s 2 R(X T X) 1 R T ) 1 (R ˆβ b) a χ 2 (m) under H 0 use s 2 = (N n) 1 i û2 i W 3 = (1/m)(R ˆβ b) T (s 2 R(X T X) 1 R T ) 1 (R ˆβ b) F (m, N n) under H 0 Hypothesis Testing 8-15

16 simple proof: W 1 is in the form of z T A 1 z where cov(z) = A, then W 1 = (A 1/2 z) T (A 1/2 z) quadratic form of standard Gaussian vector use the result on page 3-51: quadratic form of Gaussian is Chi-square W 2 = (σ 2 /s 2 )W 1 and plim(σ 2 /s 2 ) = 1, so W 2 converges to a Chi-square asymptotically (use Tranformation theorem on page 4-15) we can write W 3 as a ratio between two scaled Chi-square RVs W 3 = W 1/m s 2 /σ 2 = W 1 /m ((N n)s 2 /σ 2 )/(N n) Hypothesis Testing 8-16

17 Wald test of one restriction for a test of one restriction on linear regression model: y = Xβ + u, u N (0, σ 2 I) homoskedasticity and X is deterministic the hypotheses are H 0 : a T β b = 0, H 1 : a T β b 0 where a R n and b R for the LS estimate, it s easy to show that a T ˆβ b is Gaussian with E[a T ˆβ b] = 0, cov(a T ˆβ b) = a T σ 2 (X T X) 1 a under H 0 Hypothesis Testing 8-17

18 therefore, we can propose two Wald statistics Wald z-test statistic: W 4 = a T ˆβ b at σ 2 (X T X) 1 a N (0, 1) Wald t-test statistic: use s 2 = RSS/(N n) W 5 = a T ˆβ b at s 2 (X T X) 1 a t N n we can write W 5 a ratio of standard normal to sqrt of scaled Chi-square: W 5 = a T ˆβ b at s 2 (X T X) 1 a = T a ˆβ b a T σ 2 (X T X) 1 a (N n)s 2 σ 2 (N n) t N n Hypothesis Testing 8-18

19 Example on one exclusion restriction consider the exclusion restriction that β 1 is zero: a = (1, 0,..., 0), b = 0 suppose we use s 2 = RSS/(N n) so that Âvar( ˆβ) = s 2 (X T X) 1, Âvar( ˆβ 1 ) = ( s 2 (X T X) 1) 11 Wald test statistics for exclusion restriction are W 3 = ˆβ 2 1 Âvar( ˆβ 1 ) F (1, N n) W 5 = ˆβ 1 Âvar( ˆβ 1 ) t N n W 2 = ˆβ 2 1 Âvar( ˆβ 1 ) a χ 2 (1) if another consistent s 2 is used Hypothesis Testing 8-19

20 Nonlinear hypotheses consider hypothesis tests of m restriction that are nonlinear in θ let θ R n and r(θ) : R n R m be restriction function the null and alternative hypotheses for a two-sided tests are H 0 : r(θ ) = 0, H 1 : r(θ ) 0 examples: r(θ) = θ 2 = 0 or r(θ) = θ 1 θ 2 1 = 0 assumptions: the Jacobian matrix of r: R(θ) = Dr(θ) R m n is full rank m at θ parameters are not at the boundary of Θ under H 0, e.g., we rule out H 0 : θ 1 = 0 if the model requires θ 1 0 Hypothesis Testing 8-20

21 Wald test statistic for nonlinear restriction intuition: obtain ˆθ w/o imposing restrictions and see if r(ˆθ) 0 the Wald test statistic W = r(ˆθ) T [R(ˆθ)Âvar(ˆθ)R(ˆθ) T ] 1 r(ˆθ) is asymptotically χ 2 (m) distributed under H 0 two equivalent conditions in testing: H 0 is rejected against H 1 at significance level α if W > χ 2 α(m) H 0 is rejected at level α if the p-value: P (χ 2 (m) > W ) < α Hypothesis Testing 8-21

22 Example of nonlinear restriction let θ R n and consider a test of single nonlinear restriction H 0 : r(θ) = θ 1 /θ 2 1 = 0 then R(θ) R 1 n and given by R(θ) = [ 1/θ 2 θ 1 /θ ] let a ij be (i, j) entry of Âvar(ˆθ) W = ( ) 2 θ1 1 1 [ θ θ 2 2 θ 1 θ ] a 11 a 12 a 21 a θ 2 θ 1 θ = [θ 2 (θ 1 θ 2 )] 2 ( θ 2 2a 11 2θ 1 θ 2 a 12 + θ 2 1a 22 ) 1 a χ 2 (1) (θ is evaluated at ˆθ and the sample size N is hidden in a ij ) Hypothesis Testing 8-22

23 Derivation of the Wald statistic assumption: ˆθ has a normal limit distribution: N(ˆθ θ ) d N (0, P ) proof: starting from the first-order Taylor expansion of r under H 0 expand r around θ r(ˆθ) = r(θ ) + r(ζ)(ˆθ θ ), ζ is between ˆθ and θ write N(r(ˆθ) r(θ )) = R(ζ) N(ˆθ θ ) and note that R(ζ) p R(θ ), N(ˆθ θ ) d N (0, P ) (use that R is continuous, apply Slustky and sandwich theorems) Hypothesis Testing 8-23

24 by Product Limit Normal Rule on page 4-16 N(r(ˆθ) r(θ )) d N (0, R(θ )P R(θ ) T ) under H 0 : r(θ ) = 0 and use Avar(ˆθ) = P/N, we can write r(ˆθ) d N (0, R(θ ) Avar(ˆθ)R(θ ) T ) a quadratic form of standard Gaussian is a chi-square (on page 3-51) r(ˆθ) T [ R(θ ) Avar(ˆθ)R(θ ) T ] 1 r(ˆθ) a χ 2 (m) the Wald test statistic is obtained by using estimates of R(θ ) and Avar(ˆθ) R(ˆθ), Âvar(ˆθ) = ˆP /N Hypothesis Testing 8-24

25 Likelihood-based tests hypothesis testings when the likelihood function is known Wald test likelihood ratio (LR) test Lagrange multiplier (or score) test we denote L(θ) = f(y 1,..., y N x 1,..., x N, θ) likelihood function r(θ) : R n R m restriction function with H 0 : r(θ) = 0 ˆθ u : unrestricted MLE which maximizes L ˆθ r : restricted MLE which maximizes the Lagrangian log L(θ) λ T r(θ) Hypothesis Testing 8-25

26 Likelihood ratio test idea: if H 0 is true, the unconstrained and constrained maximization of log L should be the same it can be shown that the likelihood ratio test statistic: LR = 2[log L(ˆθ r ) log L(ˆθ u )] is asymptotically chi-square distributed under H 0 with degree of freedom m if H 0 is true, r(ˆθ u ) should be close to zero note that log L(ˆθ u ) is always greater than log L(ˆθ r ) LR test requires both ˆθ u and ˆθ r m is the number of restriction equations Hypothesis Testing 8-26

27 Wald test idea: if H 0 is true, ˆθ u should satisfy r(ˆθ u ) 0 specifically for MLE, the estimate covariance satisfies CR bound and IM equality: Avar(ˆθ u ) = I N (θ) 1 = ( E[ 2 log L(θ )] ) 1 P/N this leads to the Wald test statistic W = r(ˆθ u ) T [ R(ˆθ u )Âvar(ˆθ u )R(ˆθ u ) T ] 1 r(ˆθu ) a χ 2 (m) where Âvar(ˆθ u ) is an estimated asymptotic covariance of ˆθ u the advantage over LR test is that only ˆθ u is required Hypothesis Testing 8-27

28 Lagrange multiplier (or score) test ideas: we know that log L(ˆθ u ) = 0 (because it s an unconstrained maximization) if H 0 is true, then maximum should also occur at ˆθ r : log L(ˆθ r ) 0 LM test is called score test because log L(θ) is the score vector maximizing the Lagrangian: log L(θ) λ T r(θ) implies that log L(ˆθ r ) = r(ˆθ r ) T λ tests based on λ are equivalent to tests based on log L(ˆθ r ) because we assume r(θ) to be full rank Hypothesis Testing 8-28

29 the LM test requires the asymptotic distribution of log L(ˆθ r ) note that the asymptotic covariance of log L(ˆθ r ) is the information matrix this leads to the Lagrange multiplier test or score test statistic LM = ( log L(ˆθ r )) T [I N (ˆθ r )] 1 ( log L(ˆθ r )) which is asymptotically chi-square with m degree of freedoms Hypothesis Testing 8-29

30 Graphical interpretation of loglikelihood based tests Likelihood ratio Wald test checks if r(ˆθ u ) 0 0 Lagrange multiplier Wald LR test checks the difference between log L(ˆθ u ) and log L(ˆθ r ) LM test checks that the slope of loglikelihood at the restricted estimator should be near zero W.H. Greene, Econometric Analysis, Prentice Hall, 2008 Hypothesis Testing 8-30

31 Gaussian example consider i.i.d. example with y i N (µ, 1) with hypothesis H 0 : µ = µ 0, (µ 0 is just a constant value, and given) therefore, ˆµ u = ȳ and ˆµ r = µ 0 (restricted solution) log L(µ) = (N/2) log 2π (1/2) i (y i µ) 2 µ log L(µ) = i (y i µ) LR test: with some algebra, we can write that LR = 2[log L(ȳ) log L(µ 0 )] = N(ȳ µ 0 ) 2 Wald test: we check if ȳ µ 0 0 Hypothesis Testing 8-31

32 under H 0, the true mean of y i is µ 0, so E[ȳ] = µ 0 and var(ȳ) = 1/N hence, (ȳ µ 0 ) N (0, 1/N) W = (ȳ µ 0 )(1/N) 1 (ȳ µ 0 ) = N(ȳ µ 0 ) 2 LM test: check if log L(µ 0 ) = N(ȳ µ 0 ) 0 log L(µ 0 ) = i (y i µ 0 ) = N(ȳ µ 0 ) LM is just a rescaling of (ȳ µ 0 ), so LM = W in conclusion, the three statistics are equivalent aysmptotically but they differ in finite samples LM = W = LR Hypothesis Testing 8-32

33 MATLAB example the example is based on estimating the mean of a Gaussian model N = 50; mu = 2; y = mu + randn(n,1); ybar = mean(y); mu0 = mu; dof = 1; % number of restrictions % Wald rw = ybar-mu0; Rw = 1; EstCov = 1/N; [h,pvalue,stat,cvalue] = waldtest(rw,rw,estcov) % LR ull = -(1/2)*sum((y-ybar).^2);rLL = -(1/2)*sum((y-mu0).^2); [h,pvalue] = lratiotest(ull,rll,dof) % LM score = N*(ybar-mu0);EstCov = 1/N; % I_N(\theta) = N [h,pvalue] = lmtest(score,estcov,dof) Hypothesis Testing 8-33

34 the three tests are the same and the result is ybar = h = 0 pvalue = stat = cvalue = the p-value is greater than α = 0.05 (default value), so H 0 is accepted Hypothesis Testing 8-34

35 Poisson regression example consider the log-likelihood function in the Poisson regression model log L(β) = N [ e xt i β + y i x T i β log y i!] i=1 suppose β = (β 1, β 2 ) and H 0 : r(β) = β 2 = 0 the first and second derivatives of log L(β) are log L(β) = i (y i e xt i β )x i, 2 log L(β) = i e xt i β x i x T i unrestricted MLE, ˆβ u = ( ˆβ u1, ˆβ u2 ), satisfies log L(β) = 0 restricted MLE, ˆβ r = ( ˆβ r1, 0) where ˆβ r1 solves i (y i e xt i1 β 1)x i1 = 0 Hypothesis Testing 8-35

36 all the there statistics can be derived as LR test: calculate the fitted log-likelihood of ˆβ u and ˆβ r Wald test: compute the asympototic covariance of ˆβ u and its estimate Avar( ˆβ u ) = I N (β) 1 = E[ 2 log L(β)] 1 = ( E[ i e xt i β x i x T i ] ) 1 Âvar( ˆβ u ) = ( e xt i ˆβ u x i x T i i ) 1 from r(β) = ˆβ 2 and R(β) = [ 0 I ], Wald statistic is W = ˆβ T u2 (Âvar( ˆβu ) 22 ) 1 ˆβu2 Hypothesis Testing 8-36

37 where Âvar( ˆβ u ) 22 denotes the (2, 2) block of Âvar( ˆβ u ) LM test: it is based on log L( ˆβ r ) log L( ˆβ r ) = i (y i e xt i ˆβ r )x i = i x i û i where û i = y i e xt i1 ˆβ r1 the LM statistic is LM = [ ] T [ x i û i i e xt i1 ˆβ r1 x i x T i ] 1 [ ] x i û i i i some further simplification is possible since i x i1û i = 0 (from first-order condition) LM test here is based on the correlation between the omitted regressors and the residual, û Hypothesis Testing 8-37

38 MATLAB example data generation and solve for unrestricted and restricted MLE estimates % Data generation beta = [1 0] ; % The true value x = randn(n,2); y = zeros(n,1); lambda = zeros(n,1); for k=1:n, lambda(k) = exp(x(k,:)*beta); y(k) = poissrnd(lambda(k)); % generate samples of y end % minimization of -Loglikelihood function (change the sign of LogL) neglogfun -sum(-exp( sum(x.*repmat(beta,n,1),2) )... +y.*sum(x.*repmat(beta,n,1),2) ); beta0 = [2 2] ; % initial value [beta_u,ulogl] = fminunc(neglogfun,beta0); ulogl = -ulogl; % change back the sign of LogL Hypothesis Testing 8-38

39 % solving for restricted MLE neglogfun_r -sum(-exp( sum(x(:,1).*repmat(beta(1),n,1),2))... +y.*sum(x(:,1).*repmat(beta(1),n,1),2) ); % when beta2 = 0 [beta_r1,rlogl] = fminunc(neglogfun_r,beta0(1)) rlogl = -rlogl; beta_r = [beta_r1 0] ; The two estimates are beta_u = beta_r = Hypothesis Testing 8-39

40 Wald test TMP = 0; for ii=1:n, TMP = TMP + exp(x(ii,:)*beta_u)*x(ii,:) *x(ii,:); end rw = beta_u(2); Rw = [0 1]; EstCovw = TMP\eye(2) [h,pvalue,stat,cvalue] = waldtest(rw,rw,estcovw) h = 0 pvalue = stat = cvalue = accept H 0 since p-value is greater than α = 0.05 Hypothesis Testing 8-40

41 LR test LR = 2*(uLogL - rlogl) [h,pvalue,stat,cvalue] = lratiotest(ulogl,rlogl,dof) LR = h = 0 pvalue = stat = cvalue = Hypothesis Testing 8-41

42 LM test uhat = y - exp(sum(x.*repmat(beta_r,n,1),2)) ; scorei = x.*repmat(uhat,1,2) ; scorei = scorei ; score = sum(scorei,2); % expect of outer product of gradient of LogL EstCovlm1 = scorei*scorei ; EstCovlm2 = 0; % expectation of Hessian of LogL for ii=1:n, EstCovlm2 = EstCovlm2 + exp(x(ii,:)*beta_r)*x(ii,:) *x(ii,:); end % note that EstCovlm1 and EstCovlm2 should be close to each other % choose to use EstCovlm2 EstCovlm = EstCovlm2\eye(2); % covariance matrix of parameters [h,pvalue,stat,cvalue] = lmtest(score,estcovlm,dof) Hypothesis Testing 8-42

43 h = 0 pvalue = stat = cvalue = all the three tests agree what we should accept H 0 since p-value is greater than α Hypothesis Testing 8-43

44 if we change the true value to β = (1, 1) then ˆβ u = (1.0954, ), ˆβr = (1.3942, 0) we found that all the three tests reject H 0, i.e., β 2 is not close to zero Hypothesis Testing 8-44

45 Summary the three tests are asymptotically equivalent under H 0 but they can behave rather differently in a small sample LR test requires calculation of both restricted and unrestricted estimators Wald test requires only unrestricted estimator LM test requires only restricted estimator the choice among them typically depends on the ease of computation Hypothesis Testing 8-45

46 Hypothesis Testing introduction Wald test likelihood-based tests significance test for linear regression Hypothesis Testing 8-46

47 Recap of linear regression a linear regression model is y = Xβ + u homoskedasticity assumption: u i has the same variance for all i, given by σ 2 prediction (fitted) error: û := ŷ y = X ˆβ y residual sum of squares: RSS = û 2 2 a consistent estimate of σ 2 : s 2 = RSS/(N n) (N n)s 2 χ 2 (N n) square root of s 2 is called standard error of the regression Avar( ˆβ) = s 2 (X T X) 1 (can replace s 2 by any consistent ˆσ 2 ) Hypothesis Testing 8-47

48 Common tests for linear regression testing a hypothesis about a coefficient we can use both t and F statistics testing using the fit of the regression H 0 : β k = 0 VS H 1 : β k 0 H 0 : reduced model VS H 1 : full model if H 0 were true, the reduced model (β k = 0) would lead to smaller prediction error than that of the full model (β k 0) Hypothesis Testing 8-48

49 Testing a hypothesis about a coefficient statistics for testing hypotheses: H 0 : β k = 0 VS H 1 : β k 0 ˆβ k t N n s 2 ((X T X) 1 ) kk ( ˆβ k ) s 2 F 1,N n 2 ((X T X) 1 ) kk the above statistics are Wald statistics derived on page 8-17 through 8-19 the term s 2 ((X T X) 1 ) kk is referred to standard error of the coefficient the expression of SE can be simplified or derived in many ways (please check) e.g. R use t-statistic (two-tail test) Hypothesis Testing 8-49

50 Testing using the fit of the regression hypotheses are based on the fitting quality of reduced/full models H 0 : reduced model VS H 1 : full model reduced model: β k = 0 and full model: β k 0 the F -statistic used in this test (RSS R RSS F ) RSS F /(N n) F (1, N n) RSS R and RSS F are the residual sum squares of reduced and full models RSS R cannot be smaller than RSS F, so if H 0 were true, then the F statistic would be zero e.g. fitlm in MATLAB use this F statistic, or in ANOVA table Hypothesis Testing 8-50

51 MATLAB example perform t-test using α = 0.05 and the true parameter is β = (1, 0, 1, 0.5) realization 1: N = 100 >> [btrue b SE pvalue2side] = ˆβ is close to β it s not clear if ˆβ 2 is zero but the test decides ˆβ 2 = 0 note that all coefficients have pretty much the same SE Hypothesis Testing 8-51

52 realization 2: N = 10 >> [btrue b SE pvalue2side] = realization 3: N = 10 >> [btrue b SE pvalue2side] = some of ˆβ is close to the true value but some is not the test 2 decides ˆβ 2 and ˆβ 4 are zero while the test 3 decides all β are zero the sample size N affects type II error (fails to reject H 0 ) and we get different results from different data sets Hypothesis Testing 8-52

53 Summary common tests are available in many statistical softwares, e.g, minitab, lm in R, fitlm in MATLAB, one should use with care and interpret results correctly an estimator is random; one cannot trust its value calculated based on a data set examining statistical properties of an estimator is preferred Hypothesis Testing 8-53

54 Chapter 12 in References J.M. Wooldridge, Econometric Analysis of Cross Section and Panel Data, the MIT press, 2010 Chapter 5 in A.C. Cameron and P.K. Trivedi, Microeconometircs: Methods and Applications, Cambridge, 2005 Chapter 5, 11 and 16 in W.H. Greene, Econometric Analysis, Prentice Hall, 2008 Review of Basic Statistics (online course) Stat 501 (online course) Hypothesis Testing 8-54

Introduction Large Sample Testing Composite Hypotheses. Hypothesis Testing. Daniel Schmierer Econ 312. March 30, 2007

Introduction Large Sample Testing Composite Hypotheses. Hypothesis Testing. Daniel Schmierer Econ 312. March 30, 2007 Hypothesis Testing Daniel Schmierer Econ 312 March 30, 2007 Basics Parameter of interest: θ Θ Structure of the test: H 0 : θ Θ 0 H 1 : θ Θ 1 for some sets Θ 0, Θ 1 Θ where Θ 0 Θ 1 = (often Θ 1 = Θ Θ 0